Solving Multivariable Limit: Does it Exist?

[Petrus]

Hello MHB,

$$\lim_{(x,y)->(0,0)} \frac{6x^3y}{2x^4+x^4}$$
I did easy solve that the limit do not exist by $$(0,t)=0$$, $$(t,0)=0$$, $$(t,t)=\frac{6}{3}$$
but I wanted Also to solve this by polar cordinate so we got
$$\lim_{r->0}\frac{6\cos(\theta)\sin(\theta)}{2\cos(\theta)+ \sin(\theta)}$$
so My question is what can I say to show this limit Will never exist.
My argument: the top Will never be same for $$\theta$$ and bottom Will never be equal to zero. So the limit does not exist and this does not sound like a argument for me..

Regards,
$$|\pi\rangle$$

 

[topsquark]

Hello MHB,

$$\lim_{(x,y)->(0,0)} \frac{6x^3y}{2x^4+x^4}$$
I did easy solve that the limit do not exist by $$(0,t)=0$$, $$(t,0)=0$$, $$(t,t)=\frac{6}{3}$$
but I wanted Also to solve this by polar cordinate so we got
$$\lim_{r->0}\frac{6\cos(\theta)\sin(\theta)}{2\cos(\theta)+ \sin(\theta)}$$
so My question is what can I say to show this limit Will never exist.
My argument: the top Will never be same for $$\theta$$ and bottom Will never be equal to zero. So the limit does not exist and this does not sound like a argument for me..

Regards,
$$|\pi\rangle$$

Perhaps this will help: As r approaches 0, theta will become undefined.

-Dan

 

[Petrus]

Perhaps this will help: As r approaches 0, theta will become undefined.

-Dan

Ofcourse that's true.. I see I Was only confusing myself.. I only need to argument for like $$\lim_{r->0}\frac{r\cos(\theta)}{\cos(\theta)+\sin(\theta)}$$ that $$\sin(\theta)+cos(\theta) \neq 0$$ while this one is OBVIOUS that it Will not exist!
thanks For the help!

Regards,
$$|\pi\rangle$$

 

[Fantini]

Perhaps this will help: As r approaches 0, theta will become undefined.

-Dan

Not really. As Petrus found, what we have is that

$$\lim_{r \to 0^+} \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 + \sin^4 \theta} = \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 + \sin^4 \theta},$$

since it is no longer dependent on $r$. The point is: this is now completely dependent on $\theta$. Therefore, taking $\theta = 0$ we get the zero answer while taking $\theta = \pi/4$ we get another nonzero answer. Hence the limit does not exist.

Ofcourse that's true.. I see I Was only confusing myself.. I only need to argument for like $$\lim_{r->0}\frac{r\cos(\theta)}{\cos(\theta)+\sin(\theta)}$$ that $$\sin(\theta)+cos(\theta) \neq 0$$ while this one is OBVIOUS that it Will not exist!
thanks For the help!

Regards,
$$|\pi\rangle$$

I do not follow your reasoning. Even though it will not exist, what does it have to do with the fact that $\sin \theta + \cos \theta \neq 0$? Do not forget the exponents. The transformation to polar coordinates is $x = r \cos \theta$ and $y = r \sin \theta$, consequently

$$\frac{6x^3 y}{2x^4 + y^4} = \frac{6(r \cos \theta)^3 (r \sin \theta}{2(r \cos \theta)^4 + (r \sin \theta)^4} = \frac{r^4 6 \cos^3 \theta \sin \theta}{r^4 (2 \cos^4 \theta + \sin^4 \theta)} = \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 \theta+\sin^4 \theta}.$$

In my humble opinion your first approach is best and less confusing: two lines approaching the origin give two different limits and therefore it does not exist. Polar coordinates can be deceiving. What do you think will be the limit of

$$\lim_{(x,y) \to (0,0)} \frac{2x^2y}{x^4 +y^2}?$$

 

[Petrus]

Not really. As Petrus found, what we have is that

$$\lim_{r \to 0^+} \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 + \sin^4 \theta} = \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 + \sin^4 \theta},$$

since it is no longer dependent on $r$. The point is: this is now completely dependent on $\theta$. Therefore, taking $\theta = 0$ we get the zero answer while taking $\theta = \pi/4$ we get another nonzero answer. Hence the limit does not exist. I do not follow your reasoning. Even though it will not exist, what does it have to do with the fact that $\sin \theta + \cos \theta \neq 0$? Do not forget the exponents. The transformation to polar coordinates is $x = r \cos \theta$ and $y = r \sin \theta$, consequently

$$\frac{6x^3 y}{2x^4 + y^4} = \frac{6(r \cos \theta)^3 (r \sin \theta}{2(r \cos \theta)^4 + (r \sin \theta)^4} = \frac{r^4 6 \cos^3 \theta \sin \theta}{r^4 (2 \cos^4 \theta + \sin^4 \theta)} = \frac{6 \cos^3 \theta \sin \theta}{2 \cos^4 \theta+\sin^4 \theta}.$$

In my humble opinion your first approach is best and less confusing: two lines approaching the origin give two different limits and therefore it does not exist. Polar coordinates can be deceiving. What do you think will be the limit of

$$\lim_{(x,y) \to (0,0)} \frac{2x^2y}{x^4 +y^2}?$$

Hello,
after some simplify I get
$$\lim_{r->0}\frac{2r\cos(\theta)\sin( \theta)}{r^2\cos^2(\theta)+\sin^2(\theta)}$$
the top Will always go to zero but the bottom Will go to zero if $$\theta=0$$ so the limit does not exist?

Regards,
$$|\pi\rangle$$

 

[Fantini]

You can't think as the quantities varying independently. Regardless of what the bottom goes, the whole fraction tends to zero. Considering the case $\theta = 0$ is simply saying $y=0$, which is in agreeing with the limit we have calculated.

However, consider the curve $y=x^2$. Approaching the origin with this gives us

$$\lim_{x \to 0} \frac{2x^2 (x^2)}{x^4 + (x^2)^2} = \lim_{x \to 0} \frac{2x^4}{2x^4} = 1.$$

Even though we have found the limit as zero when considering all straight lines through the origin, as we considered a parabola we've found a different value. Therefore, the limit does not exist.

This is a good example to illustrate two things: first, it is not enough to consider all straight lines; second, polar coordinates can deceive us. :)

Cheers!

 

[Petrus]

I just find something intressting I think..

When I calculate it and facit in My book says does not exist while WA says it should be zero..?

Regards,
$$|\pi\rangle$$

 

[I like Serena]

I just find something intressting I think..

When I calculate it and facit in My book says does not exist while WA says it should be zero..?

Regards,
$$|\pi\rangle$$

Shows that W|A is a very helpful calculator.
As far as straight forward calculations go, you can trust it blindly.
But where special cases are concerned, you'd better be careful...